Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
pair<int, int> p[200010];
long long s[200010];
int n;
long long det(long long xs, long long ys, long long xt, long long yt) {
  return xs * yt - xt * ys;
}
int gl(int yp, int ym, int xp, int xm, int i, double &t) {
  t = 0;
  long long dr = det(xp - xm, yp - ym, p[i].second - xm, -ym);
  long long dl = det(xp - xm, yp - ym, p[i].first - xm, -ym);
  if (dl <= 0) return 1;
  if (dr <= 0) {
    t = 1. * dr / (ym - yp);
  }
  return 0;
}
int fl(int yp, int ym, int xp, int xm, double &t) {
  int lo = 0, hi = n;
  if (gl(yp, ym, xp, xm, 0, t)) return 0;
  while (lo + 1 < hi) {
    int mid = (lo + hi) / 2;
    if (gl(yp, ym, xp, xm, mid, t)) {
      hi = mid;
    } else {
      lo = mid;
    }
  }
  gl(yp, ym, xp, xm, lo, t);
  return lo + 1;
}
int main() {
  int sy, a, b;
  scanf("%d%d%d%d", &sy, &a, &b, &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &p[i].first, &p[i].second);
    s[i + 1] = s[i] + p[i].second - p[i].first;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    double tl, tr;
    int l = fl(y, sy, x, a, tl);
    int r = fl(y, sy, x, b, tr);
    printf("%.12f\n", (tl - tr + s[r] - s[l]) / y * (y - sy));
  }
  return 0;
}